3.17.21 \(\int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx\) [1621]

3.17.21.1 Optimal result

Integrand size = 22, antiderivative size = 54 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {2401}{5324 (1-2 x)}-\frac {1}{30250 (3+5 x)^2}-\frac {136}{166375 (3+5 x)}+\frac {9261 \log (1-2 x)}{58564}+\frac {7074 \log (3+5 x)}{1830125} \]

output

2401/5324/(1-2*x)-1/30250/(3+5*x)^2-136/166375/(3+5*x)+9261/58564*ln(1-2*x 
)+7074/1830125*ln(3+5*x)
 

3.17.21.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {\frac {3301375}{1-2 x}-\frac {242}{(3+5 x)^2}-\frac {5984}{3+5 x}+1157625 \log (1-2 x)+28296 \log (6+10 x)}{7320500} \]

input

Integrate[(2 + 3*x)^4/((1 - 2*x)^2*(3 + 5*x)^3),x]
 

output

(3301375/(1 - 2*x) - 242/(3 + 5*x)^2 - 5984/(3 + 5*x) + 1157625*Log[1 - 2* 
x] + 28296*Log[6 + 10*x])/7320500
 

3.17.21.3 Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^2 (5 x+3)^3} \, dx\)

\(\Big \downarrow \) 99

\(\displaystyle \int \left (\frac {7074}{366025 (5 x+3)}+\frac {136}{33275 (5 x+3)^2}+\frac {1}{3025 (5 x+3)^3}+\frac {9261}{29282 (2 x-1)}+\frac {2401}{2662 (2 x-1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2401}{5324 (1-2 x)}-\frac {136}{166375 (5 x+3)}-\frac {1}{30250 (5 x+3)^2}+\frac {9261 \log (1-2 x)}{58564}+\frac {7074 \log (5 x+3)}{1830125}\)

input

Int[(2 + 3*x)^4/((1 - 2*x)^2*(3 + 5*x)^3),x]
 

output

2401/(5324*(1 - 2*x)) - 1/(30250*(3 + 5*x)^2) - 136/(166375*(3 + 5*x)) + ( 
9261*Log[1 - 2*x])/58564 + (7074*Log[3 + 5*x])/1830125
 

3.17.21.3.1 Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.17.21.4 Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {-\frac {1501713}{133100} x^{2}-\frac {4502169}{332750} x -\frac {2699471}{665500}}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}+\frac {9261 \ln \left (-1+2 x \right )}{58564}+\frac {7074 \ln \left (3+5 x \right )}{1830125}\)	\(44\)
default	\(-\frac {1}{30250 \left (3+5 x \right )^{2}}-\frac {136}{166375 \left (3+5 x \right )}+\frac {7074 \ln \left (3+5 x \right )}{1830125}-\frac {2401}{5324 \left (-1+2 x \right )}+\frac {9261 \ln \left (-1+2 x \right )}{58564}\)	\(45\)
norman	\(\frac {-\frac {16205857}{598950} x^{2}-\frac {2699471}{119790} x^{3}-\frac {1621513}{199650} x}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}+\frac {9261 \ln \left (-1+2 x \right )}{58564}+\frac {7074 \ln \left (3+5 x \right )}{1830125}\)	\(47\)
parallelrisch	\(\frac {12733200 \ln \left (x +\frac {3}{5}\right ) x^{3}[phone] \ln \left (x -\frac {1}{2}\right ) x^{3}+8913240 \ln \left (x +\frac {3}{5}\right ) x^{2}[phone] \ln \left (x -\frac {1}{2}\right ) x^{2}-1484709050 x^{3}-3055968 \ln \left (x +\frac {3}{5}\right ) x -125023500 \ln \left (x -\frac {1}{2}\right ) x -1782644270 x^{2}-2291976 \ln \left (x +\frac {3}{5}\right )-93767625 \ln \left (x -\frac {1}{2}\right )-535099290 x}{65884500 \left (-1+2 x \right ) \left (3+5 x \right )^{2}}\)	\(93\)

input

int((2+3*x)^4/(1-2*x)^2/(3+5*x)^3,x,method=_RETURNVERBOSE)
 

output

50*(-1501713/6655000*x^2-4502169/16637500*x-2699471/33275000)/(-1+2*x)/(3+ 
5*x)^2+9261/58564*ln(-1+2*x)+7074/1830125*ln(3+5*x)
 

3.17.21.5 Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {82594215 \, x^{2} - 28296 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) - 1157625 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 99047718 \, x + 29694181}{7320500 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

input

integrate((2+3*x)^4/(1-2*x)^2/(3+5*x)^3,x, algorithm="fricas")
 

output

-1/7320500*(82594215*x^2 - 28296*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) 
 - 1157625*(50*x^3 + 35*x^2 - 12*x - 9)*log(2*x - 1) + 99047718*x + 296941 
81)/(50*x^3 + 35*x^2 - 12*x - 9)
 

3.17.21.6 Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {- 7508565 x^{2} - 9004338 x - 2699471}{33275000 x^{3} + 23292500 x^{2} - 7986000 x - 5989500} + \frac {9261 \log {\left (x - \frac {1}{2} \right )}}{58564} + \frac {7074 \log {\left (x + \frac {3}{5} \right )}}{1830125} \]

input

integrate((2+3*x)**4/(1-2*x)**2/(3+5*x)**3,x)
 

output

(-7508565*x**2 - 9004338*x - 2699471)/(33275000*x**3 + 23292500*x**2 - 798 
6000*x - 5989500) + 9261*log(x - 1/2)/58564 + 7074*log(x + 3/5)/1830125
 

3.17.21.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {7508565 \, x^{2} + 9004338 \, x + 2699471}{665500 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {7074}{1830125} \, \log \left (5 \, x + 3\right ) + \frac {9261}{58564} \, \log \left (2 \, x - 1\right ) \]

input

integrate((2+3*x)^4/(1-2*x)^2/(3+5*x)^3,x, algorithm="maxima")
 

output

-1/665500*(7508565*x^2 + 9004338*x + 2699471)/(50*x^3 + 35*x^2 - 12*x - 9) 
 + 7074/1830125*log(5*x + 3) + 9261/58564*log(2*x - 1)
 

3.17.21.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=-\frac {2401}{5324 \, {\left (2 \, x - 1\right )}} + \frac {2 \, {\left (\frac {1518}{2 \, x - 1} + 685\right )}}{366025 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} - \frac {81}{500} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) + \frac {7074}{1830125} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \]

input

integrate((2+3*x)^4/(1-2*x)^2/(3+5*x)^3,x, algorithm="giac")
 

output

-2401/5324/(2*x - 1) + 2/366025*(1518/(2*x - 1) + 685)/(11/(2*x - 1) + 5)^ 
2 - 81/500*log(1/2*abs(2*x - 1)/(2*x - 1)^2) + 7074/1830125*log(abs(-11/(2 
*x - 1) - 5))
 

3.17.21.9 Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^4}{(1-2 x)^2 (3+5 x)^3} \, dx=\frac {9261\,\ln \left (x-\frac {1}{2}\right )}{58564}+\frac {7074\,\ln \left (x+\frac {3}{5}\right )}{1830125}+\frac {\frac {1501713\,x^2}{6655000}+\frac {4502169\,x}{16637500}+\frac {2699471}{33275000}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \]

input

int((3*x + 2)^4/((2*x - 1)^2*(5*x + 3)^3),x)
 

output

(9261*log(x - 1/2))/58564 + (7074*log(x + 3/5))/1830125 + ((4502169*x)/166 
37500 + (1501713*x^2)/6655000 + 2699471/33275000)/((6*x)/25 - (7*x^2)/10 - 
 x^3 + 9/50)